since i got a bit rusty in complexe analysis i need a bit help with that one. Here my extraordinary bad solution so far:
I) for $\Bbb C$ the points of interest are $z=ni\pi$
with $$\lim_{z\to in\pi}\frac{e^{az^2}-e^{z}}{e^{2z}-1}=\lim_{z\to in\pi}\frac{2aze^{az^2}-e^z}{2e^{2z}}=\frac{2ani\pi e^{-an^2\pi^2}-e^{in\pi}}{2}$$
after using L' Hospital i come to the conclusion that it is a removable singularity for every $z=2\pi i$ wich actually can't be since using $z=\pi i$ already leads f to infinite so I wonder what my mistake is there and how to solve it. Trying the taylor sums looks way too complicated there.
II)more interesting to me is to determine singularities at ∞ in a formal clean way, what has to be mentioned there for gathering all exam points? So far i have:
$$f(\frac{1}{z})=\frac{e^{\frac{a}{z^2}}-e^{\frac{1}{z}}}{e^{\frac{2}{z}}-1}$$ for $z\not= 0$
$$\lim_{z\to 0}f(\frac{1}{z})=...=lim_{z\to 0}-\frac{2a}{z^3}e^{\frac{a-2z}{z^2}}+\frac{1}{z^2}e^{-\frac{1}{z}}$$
after using L' Hospital again and I'm going to argue that after here is a $e^{-\frac{1}{z}}$ in it multiplying $z^n$ for any $n \in \Bbb N$ is not changing the divergence wich makes ∞ a essentially singularity.
When $z=in\pi$ we have $z^2=-n^2\pi^2$ so the numerator looks like
$$\cases{e^{-an^2\pi^2}+1&\text{if $\,n\,$ is even}\\e^{-an^2\pi^2}-1&\text{if $\,n\,$ is odd}}$$ If this expression is nonzero, then $z=in\pi$ is a pole.
If $n$ is even, then $e^{-an^2\pi^2}+1=0$ if and only if $-an^2\pi^2=i(2k+1)\pi$ for some $k\in\mathbb{Z}$. Hence, if and only if $a=-i\frac{2k+1}{n^2\pi}$.
If $n$ is odd, then $e^{-an^2\pi^2}-1=0$ if and only if $-an^2\pi^2=i2k\pi$ for some $k\in\mathbb{Z}$. Hence, if and only if $a=-i\frac{2k}{n^2\pi}$.
Therefore, all singularities $z=in\pi$ except at most one are poles.
From the previous part, we know that there are $z$ with $|z|$ arbitrarily high such that $f$ assumes all values in a neighborhood of $\infty$ near $z$. It follows that $f$ must have an essential singularity at $\infty$.